# How do you solve the exponential equation 2^(x+1)=16^(x+2)?

Feb 22, 2017

Real solution:

$x = - \frac{7}{3}$

Complex solutions:

$x = - \frac{7}{3} + \frac{2 k \pi i}{3 \ln 2} \text{ }$ for any $k \in \mathbb{Z}$

#### Explanation:

If $a > 0$ then for any numbers $b , c$ we have:

${\left({a}^{b}\right)}^{c} = {a}^{b c}$

So:

${2}^{x + 1} = {16}^{x + 2} = {\left({2}^{4}\right)}^{x + 2} = {2}^{4 \left(x + 2\right)}$

The function $f \left(x\right) = {2}^{x}$ from $\mathbb{R}$ to $\left(0 , \infty\right)$ is strictly monotonic increasing and therefore one to one. So if $a , b$ are Real then:

${2}^{a} = {2}^{b} \text{ "<=>" } a = b$

So in our example, we must have:

$x + 1 = 4 \left(x + 2\right) = 4 x + 8$

Subtract $x + 8$ from both sides to find:

$- 7 = 3 x$

Divide both sides by $3$ and transpose to get:

$x = - \frac{7}{3}$

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Complex solutions

Note that:

${e}^{2 k \pi i} = 1$

for any integer value of $k$

Hence:

${2}^{\frac{2 k \pi i}{\ln} 2} = {e}^{\ln 2 \cdot \left(\frac{2 k \pi i}{\ln} 2\right)} = {e}^{2 k \pi i} = 1$

Hence for Complex values of $a , b$ we have:

${2}^{a} = {2}^{b} \text{ "<=>" "a = b+(2kpii)/ln 2 " for some } k \in \mathbb{Z}$

So in our example we must have:

$4 x + 8 = x + 1 + \frac{2 k \pi i}{\ln} 2$

Subtract $x + 8$ from both sides to get:

$3 x = - 7 + \frac{2 k \pi i}{\ln} 2$

Divide both sides by $3$ to get:

$x = - \frac{7}{3} + \frac{2 k \pi i}{3 \ln 2}$